Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $y = \dfrac{q - 8}{-q + 2} \div \dfrac{-3q^2 + 9q + 210}{-2q^2 - 10q + 28} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{q - 8}{-q + 2} \times \dfrac{-2q^2 - 10q + 28}{-3q^2 + 9q + 210} $ First factor out any common factors. $y = \dfrac{q - 8}{-(q - 2)} \times \dfrac{-2(q^2 + 5q - 14)}{-3(q^2 - 3q - 70)} $ Then factor the quadratic expressions. $y = \dfrac {q - 8} {-(q - 2)} \times \dfrac {-2(q + 7)(q - 2)} {-3(q + 7)(q - 10)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {(q - 8) \times -2(q + 7)(q - 2) } {-(q - 2) \times -3(q + 7)(q - 10) } $ $y = \dfrac {-2(q + 7)(q - 2)(q - 8)} {3(q + 7)(q - 10)(q - 2)} $ Notice that $(q + 7)$ and $(q - 2)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {-2\cancel{(q + 7)}(q - 2)(q - 8)} {3\cancel{(q + 7)}(q - 10)(q - 2)} $ We are dividing by $q + 7$ , so $q + 7 \neq 0$ Therefore, $q \neq -7$ $y = \dfrac {-2\cancel{(q + 7)}\cancel{(q - 2)}(q - 8)} {3\cancel{(q + 7)}(q - 10)\cancel{(q - 2)}} $ We are dividing by $q - 2$ , so $q - 2 \neq 0$ Therefore, $q \neq 2$ $y = \dfrac {-2(q - 8)} {3(q - 10)} $ $ y = \dfrac{-2(q - 8)}{3(q - 10)}; q \neq -7; q \neq 2 $